Question Number 84702 by Power last updated on 15/Mar/20
Commented by abdomathmax last updated on 15/Mar/20
$${I}\:=\int\:\:\:\frac{{dx}}{{shx}+\mathrm{1}}\:\Rightarrow{I}=\int\:\:\frac{{dx}}{\frac{{e}^{{x}} −{e}^{−{x}} }{\mathrm{2}}+\mathrm{1}}\:=\int\frac{\mathrm{2}{dx}}{{e}^{{x}} −{e}^{−{x}} \:+\mathrm{2}} \\ $$$$=_{{e}^{{x}} ={t}} \:\:\:\:\:\int\:\:\frac{\mathrm{2}}{{t}−{t}^{−\mathrm{1}} \:+\mathrm{2}}\frac{{dt}}{{t}}\:=\int\:\frac{\mathrm{2}{dt}}{{t}^{\mathrm{2}} −\mathrm{1}+\mathrm{2}{t}} \\ $$$$=\int\:\frac{\mathrm{2}{dt}}{{t}^{\mathrm{2}} \:+\mathrm{2}{t}−\mathrm{1}}\:=\int\:\frac{\mathrm{2}{dt}}{\left({t}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}}\:=\int\:\frac{\mathrm{2}{dt}}{\left({t}+\mathrm{1}−\sqrt{\mathrm{2}}\right)\left({t}+\mathrm{1}+\sqrt{\mathrm{2}}\right)} \\ $$$$=\frac{\mathrm{2}}{\mathrm{2}\sqrt{\mathrm{2}}}\int\:\:\left(\frac{\mathrm{1}}{{t}+\mathrm{1}−\sqrt{\mathrm{2}}}−\frac{\mathrm{1}}{{t}+\mathrm{1}+\sqrt{\mathrm{2}}}\right){dt} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{ln}\mid\frac{{t}+\mathrm{1}−\sqrt{\mathrm{2}}}{{t}+\mathrm{1}+\sqrt{\mathrm{2}}}\mid\:+{C} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{ln}\mid\frac{{e}^{{x}} \:+\mathrm{1}−\sqrt{\mathrm{2}}}{{e}^{{x}} \:+\mathrm{1}+\sqrt{\mathrm{2}}}\mid\:+{C} \\ $$
Commented by Power last updated on 15/Mar/20
$$\mathrm{thanks} \\ $$
Commented by abdomathmax last updated on 15/Mar/20
$${you}\:{are}\:{welcome} \\ $$